Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. What is the limiting reactant? What is the theoretical yield of \( \mathrm{Cl}_{2} \) ? \( 4 \mathrm{HCl}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{~s}) \longrightarrow \mathrm{MnCl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{~g}) \) MnO theoretical yield:

To determine the limiting reactant and the theoretical yield of chlorine gas (\( \mathrm{Cl}_2 \)) from the reaction: \[ 4 \mathrm{HCl}(\mathrm{aq}) + \mathrm{MnO}_2(\mathrm{s}) \longrightarrow \mathrm{MnCl}_2(\mathrm{aq}) + 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) + \mathrm{Cl}_2(\mathrm{g}) \] we need to follow these steps: 1. **Identify the molar ratios** from the balanced equation. 2. **Calculate the moles** of each reactant. 3. **Determine the limiting reactant** based on the stoichiometry. 4. **Calculate the theoretical yield** of \( \mathrm{Cl}_2 \). ### Step 1: Molar Ratios From the balanced equation, we see that: - 4 moles of \( \mathrm{HCl} \) react with 1 mole of \( \mathrm{MnO}_2 \) to produce 1 mole of \( \mathrm{Cl}_2 \). ### Step 2: Calculate Moles of Reactants Assume we have the following amounts of reactants: - \( n \) moles of \( \mathrm{HCl} \) - \( m \) moles of \( \mathrm{MnO}_2 \) ### Step 3: Determine the Limiting Reactant Using the stoichiometric ratios: - For every 4 moles of \( \mathrm{HCl} \), 1 mole of \( \mathrm{MnO}_2 \) is required. - Therefore, the amount of \( \mathrm{HCl} \) needed for \( m \) moles of \( \mathrm{MnO}_2 \) is \( 4m \) moles. Now, we compare the available moles: - If \( n < 4m \), then \( \mathrm{HCl} \) is the limiting reactant. - If \( n \geq 4m \), then \( \mathrm{MnO}_2 \) is the limiting reactant. ### Step 4: Calculate Theoretical Yield of \( \mathrm{Cl}_2 \) The theoretical yield of \( \mathrm{Cl}_2 \) can be calculated based on the limiting reactant: - If \( \mathrm{HCl} \) is limiting: \[ \text{Moles of } \mathrm{Cl}_2 = \frac{n}{4} \] - If \( \mathrm{MnO}_2 \) is limiting: \[ \text{Moles of } \mathrm{Cl}_2 = m \] ### Example Calculation Let's say we have: - 10 moles of \( \mathrm{HCl} \) - 2 moles of \( \mathrm{MnO}_2 \) 1. **Determine the limiting reactant**: - Required \( \mathrm{HCl} \) for 2 moles of \( \mathrm{MnO}_2 \): \( 4 \times 2 = 8 \) moles. - Since we have 10 moles of \( \mathrm{HCl} \), \( \mathrm{MnO}_2 \) is the limiting reactant. 2. **Calculate the theoretical yield of \( \mathrm{Cl}_2 \)**: - Moles of \( \mathrm{Cl}_2 \) produced = \( 2 \) moles (since \( \mathrm{MnO}_2 \) is limiting). ### Conclusion - **Limiting Reactant**: \( \mathrm{MnO}_2 \) - **Theoretical Yield of \( \mathrm{Cl}_2 \)**: \( 2 \) moles You can adjust the amounts of \( \mathrm{HCl} \) and \( \mathrm{MnO}_2 \) based on your specific scenario to find the limiting reactant and theoretical yield accordingly.